Here is a quote from the wikipedia entry on semidirect products of groups:

There are also far-reaching generalisations in category theory. They show how to construct fibred categories from indexed categories. This is an abstract form of the outer semidirect product construction.

Do you know any reference for this general construction? I know that every "indexed category" $F : C^{op} \to \mathrm{Cat}$ gives rise to a fibration $E \to C$, namely whose $x$-fiber is precisely $F(x)$, but is this really meant here? I don't see the connection with semi-direct products yet.

Why don't you react at all to the very long answer you got?
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t.b.Dec 7 '11 at 16:29

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"Am I supposed to look here every day?" No. But you asked 3 questions since this answer was posted. I posted a comment and you posted yet another question after that comment. Hence the downvote, hoping that you will look what happened. Obviously it worked. I removed the downvote by the way.
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t.b.Dec 9 '11 at 11:38

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Oh my god. Recently the stats page has become more confusing than before. And often I work in various problems simultanously. There is no reason for you to get me doing this in a specific order. Good answers last forever, there is no reason why I have to read and check them immediately.
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Martin BrandenburgDec 9 '11 at 20:20

1 Answer
1

A semidirect product of two groups $H$ and $N$ is the same as a group
homomorphism $\varphi: H\rightarrow \mathrm{Aut}(N)$ i.e. an action of
$H$ on $N$. Groupies usually drop the $\varphi$ from the notation and
simply write $n^h$ instead of $n^{\varphi h}$. The resulting multiplication
on $H\times N$ is then given by $(h,n)(k,m) = (hk,n^km)$.

Likewise a semidirect product of two monoids $H$ and $N$ is the same as a
monoid homomorphism $\varphi: H\rightarrow \mathrm{End}(N)$. The multiplication
is given in the same way.

Now consider a functor $F: C\rightarrow \mathrm{Cat}$ (I ignore dualization,
so this correspond to an op-fibration) and suppose that

(1) $C$ is a monoid (i.e. has exactly one object $*$)

(2) The category $F(*)$ is also a monoid.

Because an endofunctor on $F(*)$ is the same as a monoid homomorphism, this
is the same as giving a monoid homomorphism from $C$ to $\mathrm{End}(F(*))$.

In fact, if you go through the Grothendieck construction for the corresponding
op-fibration, you will produce exactly the above multipilcation. The similarity
is also visible in the general case if you view $F: C\rightarrow \mathrm{Cat}$
as a category action of $C$, drop the $F$ from the notation and write
$uf$ instead of $F(f)u$ for $f:c\rightarrow c'$ and $u\in F(c)$:

For $c,c',c''\in C$, $x\in F(c)$, $x'\in F(c')$, $x''\in F(c'')$,
$f:c\rightarrow c'$, $u:F(f)x\rightarrow x'$,
$g:c'\rightarrow c''$, $u:F(g)x'\rightarrow x''$ consider the maps
$(f,u):(c,x)\rightarrow (c',x')$ and $(g,v):(c',x')\rightarrow (c'',x'')$
in the Grothendieck construction. If we write composition of maps from
left to right and apply the above mentioned notational simplification, then
the composition of $(f,u)$ and $(g,v)$ can be written as $(fg,(ug)v)$
where $(xf)g$ is the composite
$$xfg \stackrel{ug}{\rightarrow} x'g \stackrel{v}{\rightarrow} x''$$